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GrogVix [38]
3 years ago
6

Avery jumped over 15 hurdles. If these hurdles are 60% of the hurdles in the race, how many hurdles are there in all?

Mathematics
2 answers:
Andre45 [30]3 years ago
6 0
You could set up a proportion to figure out the anwer. 15 hurdles/60%=X hurdles/100%. Cross multiply, 1500=60x. Divide by 60, 25=X=total amount of hurdles. Hope this helps!
dalvyx [7]3 years ago
6 0
40/100*15

cancel 100 and 15

40/20*3

120/20

6 hurdles is 40%

6+15=21 hurdles 

there are 21 hurdles. <span />
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A survey reports that 67% of college students prefer to drink more coffee during the exams week. If we randomly select 80 colleg
Akimi4 [234]

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The probability that at most 50 say that they drink coffee during exam week is 0.166.

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The random variable <em>X</em> can be defined as the number of college students who prefer to drink more coffee during the exams week.

The probability of the random variable <em>X</em> is <em>p</em> = 0.67.

A random sample of <em>n</em> = 80 college students are selected.

The response of every students is independent of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 80 and <em>p</em> = 0.67.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

np=80\times 0.67=53.6>10\\n(1-p)=80\times (1-0.67)=26.4>10

Thus, a Normal approximation to binomial can be applied.

X\sim N(np, np(1-p))

The mean of the distribution of <em>X</em> is:

\mu=np=80\times 0.67=53.6

The standard deviation of the distribution of <em>X</em> is:

\sigma=\sqrt{np(1-p)}=\sqrt{80\times 0.67\times (1-0.67)}=4.206

A Normal distribution is a continuous distribution. So, the probability at a point cannot be computed for the Normal distribution. To compute the probability at a point we need to apply the continuity correction.

Compute the probability that at most 50 say that they drink coffee during exam week as follows:

Apply continuity correction:

P(X\leq 50)=P(X

                 =P(X

*Use a <em>z</em>-table for the probability.

Thus, the probability that at most 50 say that they drink coffee during exam week is 0.166.

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